Polynomial equation with rational numbers

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Polynomial equation with rational numbers

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Source: Romanian TST 1979 day 2 P1

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#1 h Apr 15, 2023, 2:28 PM

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Determine the polynomial $P\in \mathbb{R}[x]$ for which there exists $n\in \mathbb{Z}_{>0}$ such that for all $x\in \mathbb{Q}$ we have: $P\left(x+\frac1n\right)+P\left(x-\frac1n\right)=2P(x).$
Dumitru Bușneag

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#2 h Apr 15, 2023, 4:45 PM

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Miquel-point wrote:

Determine the polynomial $P\in \mathbb{R}[x]$ for which there exists $n\in \mathbb{Z}_{>0}$ such that for all $x\in \mathbb{Q}$ we have: $P\left(x+\frac1n\right)+P\left(x-\frac1n\right)=2P(x).$

If $P(x)$ fits, $P'(x)$ fits too.
Easy to see that no degree $2$ polynomial can fit and so no polynomial with degree $\ge 2$

And since any $\boxed{ax+b}$ fits, these are all the polynomials.

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#3 h Yesterday at 8:23 AM

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Easier - just note that $P(x) - P\left(x + \frac{1}{n}\right)$ is constant, so (by comparison of coefficients) $P$ is linear.

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